Problem: Solve for $x$ and $y$ using substitution. ${4x+y = 7}$ ${x = -y-2}$
Since $x$ has already been solved for, substitute $-y-2$ for $x$ in the first equation. ${4}{(-y-2)}{+ y = 7}$ Simplify and solve for $y$ $-4y-8 + y = 7$ $-3y-8 = 7$ $-3y-8{+8} = 7{+8}$ $-3y = 15$ $\dfrac{-3y}{{-3}} = \dfrac{15}{{-3}}$ ${y = -5}$ Now that you know ${y = -5}$ , plug it back into $\thinspace {x = -y-2}\thinspace$ to find $x$ ${x = -}{(-5)}{ - 2}$ $x = 5 - 2$ ${x = 3}$ You can also plug ${y = -5}$ into $\thinspace {4x+y = 7}\thinspace$ and get the same answer for $x$ : ${4x + }{(-5)}{= 7}$ ${x = 3}$